Harm osc, anharmonic perturbation

  • Thread starter WarnK
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  • #1
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Homework Statement


Particle bound by
[tex] V = \frac{1}{2} m \omega^2 x^2 - a x^3 [/tex]
for small x. Show that the mean position of the particle changes with the energy of the eigenstates when [tex]a[/tex] is small, so first order perturbation theory works.

Homework Equations


For the harmonic oscillator
[tex] x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a) [/tex]


The Attempt at a Solution


That x^3 perturbation will give an odd number of creation/destruction operators, so there's no shift in energy eigenvalues to first order in perturbation theory. But how does that help answering the question?
 

Answers and Replies

  • #2
Dr Transport
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There is a first order transition, but it is between different energy states.
 
  • #3
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But how does that help answering the question?
 
  • #4
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Noone have any ideas on this?
 
  • #5
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I'm not going to give up on this! The question asks to calculate the mean position of the particle, so if I do that:
[tex]<x> = \sqrt{\frac{\hbar}{2m\omega}}<n|a^{\dagger}+a|n> = 0 [/tex]
That's just zero, haveing nothing to do with energy eigernstates or size of a?

Sure I could calculate the perturbated energy to second order, getting terms like
[tex] a^2 \big( \frac{\hbar}{2m\omega} \big)^3 <n|a^{\dagger} aa \frac{1}{E_n-H_0} a^{\dagger} a a^{\dagger} |n> = -n(n+1)^2 \frac{\hbar^2 a^2}{8m^3 \omega^4}[/tex]
and so on, but I don't see how that has anyhting to do with the question asked.
 
  • #6
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Shouldn't you consider the expanded eigenkets?

[tex] | n > = |n^{(0)}>+\lambda |n^{(1)}> + \lambda^2 |n^{(2)}> + \dots[/tex]

Then you get to first order something like:

[tex] <x> = < n^{(0)} | x | n^{(0)} > + \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2) = \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2)[/tex]

Could that be the correct approach?
 
Last edited:
  • #7
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I think that you dont really have to solve for the corrected energy's. You need the new eigenkets, and then just solve for the expectation value, <x>, by taking the inner product of x and the eigenkets.
 

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